【POJ 2318】TOYS

1.​​题目链接​​。题目大意就是有一个矩形的盒子，里面被n条线段分割成了几个不同的区域，然后给出m个点的左边，输出一下每个区域里面格子落了多少个点。

2.分析:几何的基础题了

                   

如果p在向量AB的左边，那么PA叉乘PB是一个负数，右手定则轻松解释。然后这个是具有单调性的，所以我们直接二分这个位置就行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#pragma warning(disable:4996)
using namespace std;
struct Point
{
  int  x, y;
  Point() {};
  Point(int  x_, int  y_) :x(x_), y(y_) {};
  Point operator-(const Point&b)const
  {
    return Point(x - b.x, y - b.y);
  }
  int   operator*(const Point&b)const
  {
    return x * b.x + y * b.y;
  }
  int  operator^(const Point&b)const
  {
    return x * b.y - y * b.x;
  }
};
struct Line
{
  Point s, e;
  Line() {};
  Line(Point _s, Point _e ): s(_s), e(_e){};
};
int xmult(Point p0, Point p1, Point p2)
{
  return (p1 - p0) ^ (p2 - p0);
}
const int MAXN = 5050;
Line line[MAXN];
int ans[MAXN];
int main()
{
  int n, m;
  int x1, y1, x2, y2;
  bool f = 1;
  while (~scanf("%d", &n) && n)
  {
    if (f)f = 0;
    else puts("");
    scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
    int Ui, Li;
    for (int i = 0; i < n; i++)
    {
      scanf("%d%d", &Ui, &Li);
      line[i] = Line(Point(Ui, y1), Point(Li, y2));
    }
    line[n] = Line(Point(x2, y1), Point(x2, y2));
    int x, y;
    Point p;
    memset(ans, 0, sizeof(ans));
    while (m--)
    {
      scanf("%d%d", &x, &y);
      p = Point(x, y);
      int l = 0, r = n;
      int tmp;
      while (l <= r)
      {
        int mid = (l + r) >> 1;
        if (xmult(p, line[mid].s, line[mid].e) < 0)
        {
          tmp = mid;
          r = mid - 1;
        }
        else l = mid + 1;
      }
      ans[tmp]++;
    }
    for (int i = 0; i <= n; i++)
      printf("%d: %d\n", i, ans[i]);
  }
  return 0;
}